Problem: Assume that $C$ is a negatively oriented, piecewise smooth, simple, closed curve. Let $R$ be the region enclosed by $C$. Use the normal form of Green's theorem to rewrite $ \oint_C \dfrac{x}{y^2} \, dx + 4\ln(x^2y) \, dy$ as a double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \iint_R \dfrac{-4}{y} + \dfrac{1}{x^2} \, dA$ (Choice B) B $ \iint_R \dfrac{8}{x} + \dfrac{2x}{y^3} \, dA$ (Choice C) C $ \iint_R \dfrac{4}{y} - \dfrac{1}{x^2} \, dA$ (Choice D) D $ \iint_R \dfrac{-8}{x} - \dfrac{2x}{y^3} \, dA$ (Choice E) E Green's theorem is not necessarily applicable.
Solution: Assume we have a two-dimensional vector field $F(x, y) = P(x, y) \hat{\imath} + Q(x, y) \hat{\jmath}$ and a positively oriented, piecewise smooth, simple, closed curve $C$. If $R$ is the region enclosed by $C$, then the normal form of Green's theorem states: $ \oint_C -Q \, dx + P \, dy = \iint_R \left( \dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} \right) dA$ If $C$ is negatively oriented, then the line integral is equal to the negative of the double integral. [How is this equivalent to the circulation form of Green's thoeorem?] Our first step should be to confirm that the given curve is compatible with using Green's theorem. The curve $C$ does satisfy all the properties required for Green's theorem. It is also negatively oriented, so we'll need to take the negative of our result at the end. The next step is to identify the components $P$ and $Q$ of the vector field $F$ over which we're integrating. We're simplifying a line integral into a double integral, so we need to match up the $dx$ term to $-Q$ and the $dy$ term to $P$ : $\begin{aligned} & \oint_C \dfrac{x}{y^2} \, dx + 4\ln(x^2y) \, dy \\ \\ &P(x, y) = 4\ln(x^2y) \\ \\ &Q(x, y) = \dfrac{-x}{y^2} \end{aligned}$ Our final step is to find $\dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y}$. $\begin{aligned} &P_x = \dfrac{4}{x^2y} (2xy) = \dfrac{8}{x} \\ \\ &Q_y = \dfrac{2x}{y^3} \\ \\ &\dfrac{\partial P}{\partial x} + \dfrac{\partial Q}{\partial y} = \dfrac{8}{x} + \dfrac{2x}{y^3} \end{aligned}$ Because $C$ is negatively oriented, we need to use $-(P_x + Q_y)$. Putting everything together, we get this double integral: $ \iint_R \dfrac{-8}{x} - \dfrac{2x}{y^3} \, dA$